|Sep23-04, 12:45 PM||#1|
So I understand what a hermitian operator is and how if A and B are hermitian operators, then the product of AB is not necessarily Hermitian since
*Note here + is dagger
I also recognize that (AB-BA) is not Hermitian since (AB-BA)+=B+A+-A+B+
In addition, I know that any real number a is a Hermitian operator since <Psi l|a Psi n>=<aPsi l|Psi n>
Now here comes my questions.
Where A and B are both hermitian operators,
1)how do we know if something like i(AB-BA) is a hermitian operator where i is an imaginary number? How do I show that this is not a hermitian operator because I am pretty sure it is not?
2) and how would I show that (AB+BA/2) is Hermitian because I feel like it should be, but I dont know how to interchange the 2 with the A and B operators?
And if operator A corresponds to observable A, and operator B corresponds to observable B, what is a "good" (i.e.Hermitian) operator that corresponds to the physically observable product AB?
When I am dealing with two operators, I dont think I am confused on how to work with them, but when dealing with 3 I get a little iffy. Peace and love.
|Sep23-04, 01:15 PM||#2|
So if A and B do not commute, the difference between your operator and its conjugate is not 0 (because equal to the commutator divided by 2).
(Wigner's prescription, I think it is called).
It is not unique, of course ;
you can have 1/2(AB + BA) + c i [A,B] with c an arbitrary real number for example.
|Sep23-04, 01:27 PM||#3|
1) Just use the definition of "Hermitian" : A+ = A and B+ = B (they are hermitian)
so you get :
[ i(AB-BA)]+ = (-i)(B+ A+ - A+ B+) = (-i) (BA-AB) and this equals i(AB-BA)
2) Same system, man, just use the definition of hermitian operators :
(AB+BA/2)+ - (AB+BA/2) = (BA + AB/2) - AB - BA/2 = BA/2 - AB/2 = [B,A] / 2
If A and B are not commuting operators then the difference between the given operator and its conjugate is not 0 . If A and B commute then the commutator will be 0 !!
3) this is an easy one and the answer is
1/2 (AB + BA)
|Sep23-04, 01:28 PM||#4|
I did not see you already answered this question...
I apologize for that...
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